A clarification about today’s lecture:
The Oracle. My explanation for why the oracle was an easy problem was not very clear. Let me try again. Let us define
Notice that this has non-negativity constraints, and one equality constraint. Very simple constraints. Now the oracle asks: given , find me a point such that
or else tell me no such point exists. How would you solve this?
Let’s just try to maximize for . If we can make at least , then we can return that point. If we cannot reach , we return “No such point.”
OK, suppose all entries of are positive. Then I claim you might as well make non-zero in a single coordinate. Which coordinate? The entry that maximizes
Then you’ll make to satisfy . (Of course, if , you’ll return “No such point.”)
If has negative entries, then things are a more interesting. You can still ensure that will be non-zero in at most two entries. And then you have to do a simple case analysis. I’ll omit the details here; let me know if you’d like me to say more.
The next time, we’ll show that if you can solve the oracle problem you can solve LPs (up to additive error).